\newproblem{lay:1_3_25}{
   % Problem identification
	 \begin{large}
 	   \hspace{\fill}\newline
     \textbf{Lay, 1.3.25}
	 \end{large}
	 \\
   \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

   % Problem statement
   Let $A=\begin{pmatrix}1 & 0 & -4 \\ 0 & 3 & -2 \\ -2 & 6 & 3\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix} 4 \\ 1 \\ -4 \end{pmatrix}$. Denote the columns of $A$
	 as $\mathbf{a}_1$, $\mathbf{a}_2$, and $\mathbf{a}_3$, and let $W=\mathrm{Span}\{\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3\}$
	 \begin{enumerate}[a]
		  \item Is $\mathbf{b}$ in $\{\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3\}$? How many vector are in $\{\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3\}$?
			\item Is $\mathbf{b}$ in $W$? How many vectors are in $W$?
			\item Show that $\mathbf{a}_1$ is in $W$.
	 \end{enumerate}
}{
   % Solution
	 \begin{enumerate}[a]
		  \item No, $\mathbf{b}$ is not equal to any of the columns of $A$. In $\{\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3\}$ there are only 3 vectors.
			\item In $W$ there are infinite vectors. To check whether $\mathbf{b}$ is in $W$ all we have to do is to find coefficients $x_1$, $x_2$, and $x_3$ such that
						\begin{center}
							$x_1\mathbf{a}_1+x_2\mathbf{a}_2+x_3\mathbf{a}_3=\mathbf{b}$
						\end{center}
						or what is the same
						\begin{center}
							$\left(\begin{array}{c}x_1 -4x_2 \\ 3x_2 -2x_3 \\ -2x_1+6x_2+3x_3\end{array}\right)=
							\left(\begin{array}{r}4 \\ 1 \\-4\end{array}\right)$
						\end{center}
						Solving the equation system we find: $x_1=-4$, $x_2=-1$, $x_3=-2$, i.e.
						\begin{center}
							$\mathbf{b}=-4\mathbf{a}_1-\mathbf{a}_2-2\mathbf{a}_3=-4\begin{pmatrix}1\\0\\-2\end{pmatrix}-\begin{pmatrix}0\\3\\6\end{pmatrix}-
							   2\begin{pmatrix}-4\\-2\\3\end{pmatrix}=\begin{pmatrix}4\\1\\-4\end{pmatrix}$
						\end{center}
						and, consequently, $\mathbf{b}\in\mathrm{Span}\{\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3\}$.
						
			\item It is enough to observe that
						\begin{center}
							$\mathbf{a}_1=1\mathbf{a}_1+0\mathbf{a}_2+0\mathbf{a}_3$
						\end{center}
	 \end{enumerate}
}

\useproblem{lay:1_3_25}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
